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Q.

One mole of a diatomic gas undergoes a process P=P01+V/V03 where P0 and V0 are constants. The translational kinetic energy of the gas when V = V0 is given by

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a

5P0V0/4

b

3P0V0/4

c

3P0V0/2

d

5P0V0/2

answer is B.

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Detailed Solution

P=P01+V/V03=P01+V0/V03=P02∵V=V0For 1 mole, PV=RT P02V0=RT T=P0V02RTherefore, translational kinetic energy is equal to 32RT=3R2P0V02R=3P0V04
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